Following up on Analytical Solution for the Convolution of Signal with a Box Filter, I am now trying to convolve a Gaussian filter with the sine signal by hand.
My method is to use the definition of convolution and attempt to integrate: \begin{align*} \overline{\phi} &= \phi(x) * h(x) = \int_{-\infty}^{+\infty} \! \phi(x') h(x - x') \, \mathrm{d} x' \\ \Rightarrow \overline{\phi} &= \int_{-\infty}^{+\infty} \! \sin(x') \left( \frac{6}{\pi \Delta^2} \right)^{1/2} \exp(-\frac{6 (x - x')^2}{\Delta^2}) \, \mathrm{d} x' \\ \overline{\phi} &= \left( \frac{6}{\pi \Delta^2} \right)^{1/2} \int_{-\infty}^{+\infty} \! \sin(x') \exp(-\frac{6 (x - x')^2}{\Delta^2}) \, \mathrm{d} x' \\ \end{align*} At this stage, the integration seems impossible by hand (e.g., integration by parts).
I also tried moving to wave space by taking Fourier transforms and instead using multiplication:
\begin{align*} \mathscr{F}\{\phi(x)*h(x)\} &= \mathscr{F}\{\phi(x)\} \cdot \mathscr{F}\{h(x)\} \\ \end{align*}
I get the Fourier transform of sine as $\sqrt{\pi/2} i (\delta(k - 1) - \delta(k + 1)$, but am not sure how to multiply this with the exponential and how the inverse Fourier transform will work. Am I missing something clever and/or elegant? Is such a solution possible? I don't have a good understanding of the Dirac delta and how to multiply it with an exponential in wave space.